Anharmonic Oscillator by Perturbation Theory

This worked example computes the leading perturbative corrections for the quartic anharmonic oscillator

$$H = \frac{P^2}{2m} + \frac12m\omega^2X^2 + \lambda X^4, \qquad \lambda\gt0.$$

In this worked problem, $\Delta E_n^{(r)}$ denotes the physical order-$r$ contribution and therefore includes the corresponding power of $\lambda$.

The unperturbed problem is the harmonic oscillator, and the perturbation is

$$V=\lambda X^4.$$

Method Choice

Use nondegenerate time-independent perturbation theory because the one-dimensional harmonic oscillator has isolated energy levels

$$E_n^{(0)} = \hbar\omega \left(n+\frac12\right).$$

The small parameter is the size of the quartic energy in the oscillator ground-state width compared with $\hbar\omega$:

$$g = \frac{\lambda\hbar}{m^2\omega^3}.$$

Low-order perturbation theory is plausible when $g\ll1$ and the state number $n$ is not so large that the quartic term dominates.

Ladder-Operator Setup

Write

$$X = x_{\mathrm{zpf}} \left(a+a^\dagger\right), \qquad x_{\mathrm{zpf}} = \sqrt{\frac{\hbar}{2m\omega}}.$$

Then

$$X^4 = x_{\mathrm{zpf}}^4 \left(a+a^\dagger\right)^4.$$

The diagonal matrix element is

$$\langle n|X^4|n\rangle = x_{\mathrm{zpf}}^4 \left( 6n^2+6n+3 \right).$$

This can be found by normal ordering, by repeated action of $a+a^\dagger$, or by using known oscillator moments.

First-Order Energy Shift

The first-order correction is

$$\Delta E_n^{(1)} = \lambda\langle n|X^4|n\rangle.$$

Therefore

$$\Delta E_n^{(1)} = \lambda \left( \frac{\hbar}{2m\omega} \right)^2 \left( 6n^2+6n+3 \right).$$

Equivalently,

$$\Delta E_n^{(1)} = \lambda \frac{3\hbar^2}{4m^2\omega^2} \left( 2n^2+2n+1 \right).$$

For the ground state,

$$\Delta E_0^{(1)} = \frac{3\lambda\hbar^2}{4m^2\omega^2}.$$

Second-Order Ground-State Shift

For the ground state,

$$\Delta E_0^{(2)} = \sum_{r\ne0} \frac{ |\langle r|\lambda X^4|0\rangle|^2 }{ E_0^{(0)}-E_r^{(0)} }.$$

Parity implies that only even $r$ can contribute. Repeated ladder-operator action gives

$$X^4\lvert0\rangle = x_{\mathrm{zpf}}^4 \left( 3\lvert0\rangle + 6\sqrt2\,\lvert2\rangle + 2\sqrt6\,\lvert4\rangle \right).$$

Thus only $\lvert2\rangle$ and $\lvert4\rangle$ contribute to the sum:

$$\begin{aligned} \Delta E_0^{(2)} =& \lambda^2x_{\mathrm{zpf}}^8 \left[ \frac{|6\sqrt2|^2}{E_0^{(0)}-E_2^{(0)}} + \frac{|2\sqrt6|^2}{E_0^{(0)}-E_4^{(0)}} \right] \\ =& \lambda^2x_{\mathrm{zpf}}^8 \left[ \frac{72}{-2\hbar\omega} + \frac{24}{-4\hbar\omega} \right]. \end{aligned}$$

Therefore

$$\Delta E_0^{(2)} = - \frac{42\lambda^2x_{\mathrm{zpf}}^8}{\hbar\omega} = - \frac{21\lambda^2\hbar^3}{8m^4\omega^5}.$$

Result

The ground-state energy through second order is

$$E_0 = \frac12\hbar\omega + \frac{3\lambda\hbar^2}{4m^2\omega^2} - \frac{21\lambda^2\hbar^3}{8m^4\omega^5} + O(\lambda^3).$$

In terms of

$$g=\frac{\lambda\hbar}{m^2\omega^3},$$

this is

$$\frac{E_0}{\hbar\omega} = \frac12 + \frac34g - \frac{21}{8}g^2 + O(g^3).$$

Validity Discussion

The first correction is small compared with the unperturbed ground energy when

$$g\ll1.$$

The second-order term is smaller than the first when, roughly,

$$\frac{21g^2/8}{3g/4} = \frac72g \ll1.$$

This numerical estimate is not a rigorous bound, but it is a useful warning: even for the ground state, the second-order coefficient is sizable.

For high $n$, the first-order shift grows like $n^2$, while the unperturbed energy grows like $n$. A perturbative treatment valid for low states can fail for highly excited states.

Cross-Checks

The correction has the right dimensions: $\lambda X^4$ is an energy.

The first-order shift is positive for $\lambda\gt0$, as expected for a positive added potential. The second-order shift is negative because excited-state denominators satisfy

$$E_0^{(0)}-E_r^{(0)}\lt0.$$

Parity is respected: $X^4$ connects the ground state only to even oscillator states.

Common Mistakes

• Forgetting that the perturbation is $\lambda X^4$, not just $X^4$.
• Dropping off-diagonal matrix elements when computing second order.
• Missing the $\lvert4\rangle$ contribution in $X^4\lvert0\rangle$.
• Assuming a positive perturbing potential gives positive corrections at every perturbative order.
• Treating the asymptotic perturbation series as a convergent power series.

Cross-Links

• Worked Problems and Model Calculations
• Anharmonic Oscillator by Variational Methods
• Nondegenerate Perturbation Theory
• First-Order Energy Corrections
• Second-Order Energy Corrections
• Perturbation Theory for the Harmonic Oscillator
• Anharmonic Oscillator
• Small Parameters and Error Estimates
• Ladder-Operator Solution
• Matrix Diagonalization

References

• C. M. Bender and T. T. Wu, “Anharmonic oscillator,” Physical Review 184, 1231-1260, 1969.
• C. M. Bender and T. T. Wu, “Anharmonic oscillator. II. A study of perturbation theory in large order,” Physical Review D 7, 1620-1636, 1973.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Further Exercises

1. Show that

$$\langle1|X^4|1\rangle = 15x_{\mathrm{zpf}}^4.$$

Solution

Use the diagonal formula

$$\langle n|X^4|n\rangle = x_{\mathrm{zpf}}^4(6n^2+6n+3).$$

For $n=1$,

$$6+6+3=15.$$

2. Why does a cubic perturbation $\lambda X^3$ give no first-order energy shift for oscillator eigenstates?

Solution

Harmonic oscillator eigenstates have definite parity. The operator $X^3$ is odd under parity, so

$$\langle n|X^3|n\rangle=0.$$

The leading nonzero energy shift from a cubic perturbation therefore occurs at second order, assuming the perturbative setup is physically meaningful.